mysqli_result::free

mysqli_result::close

mysqli_result::free_result

mysqli_free_result

(PHP 5, PHP 7)

mysqli_result::free -- mysqli_result::close -- mysqli_result::free_result -- mysqli_free_result — Frees the memory associated with a result

说明

面向对象风格

mysqli_result::free ( void ) : void
mysqli_result::close ( void ) : void
mysqli_result::free_result ( void ) : void

过程化风格

mysqli_free_result ( mysqli_result $result ) : void

Frees the memory associated with the result.

Note:

You should always free your result with mysqli_free_result(), when your result object is not needed anymore.

参数

result

仅以过程化样式：由 mysqli_query()，mysqli_store_result() 或 mysqli_use_result()返回的结果集标识。

返回值

没有返回值。

参见

User Contributed Notes

jack_action100 at hotmail dot example dot com 20-Mar-2019 12:43
If you are STILL getting this error, even after freeing your results:
Internal SQL Bug: 2014, Commands out of sync; you can't run this command now

You may have a stored procedure in your query.   A procedure can return more than one result set, and it will always return one extra empty result set that carries some meta information on the procedure call itself, especially error information. ( source:  https://bugs.mysql.com/bug.php?id=71044 )

While calling single procedures, with one SELECT in them, using mysqli->query("CALL `stored_procedure`();"), I had to do the following to make it work between two calls:

<?php
$result->free();
$mysqli->next_result();
?>

It has no negative impact if you are not calling a stored procedure.
Vector at ionisis dot com 28-Dec-2009 05:17
If you are getting this error:
Internal SQL Bug: 2014, Commands out of sync; you can't run this command now

Then you never called mysqli_result::free(), mysqli_result::free_result(), mysqli_result::close(), or mysqli_free_result() in your script, and must call it before executing another stored procedure.
Anonymous 18-Oct-2009 03:49
Freeing the memory associated with a result means that the references returned by mysqli_fetch_object (or equivalent) are cleared. Thus if you should pass an object pointing to a database row _by reference_, every call of mysqli_free_result will discard the referenced data.